Proof.

Let G,H and Φ be as in the statement of the lemma and let g∈G and k∈Ker⁡(Φ). Then, Φ⁢(k)=eH by definition and:

Φ⁢(g∗Gk∗Gg-1)

=

Φ⁢(g)∗HΦ⁢(k)∗HΦ⁢(g-1)

=

Φ⁢(g)∗H(eH)∗HΦ⁢(g-1)

=

Φ⁢(g)∗HΦ⁢(g-1)

=

Φ⁢(g)∗HΦ⁢(g)-1

=

eH,

where we have used several times the properties of group homomorphisms and the properties of the identity element eH. Thus, Φ⁢(g⁢k⁢g-1)=eH and g⁢k⁢g-1∈G is also an element of the kernel of Φ. Since g∈G and k∈Ker⁡(Φ) were arbitrary, it follows that Ker⁡(Φ) is normal in G.
∎

Lemma 2.

Let G be a group and let K be a normal subgroup of G. Then there exists a group homomorphism Φ:G→H, for some group H, such that the kernel of Φ is precisely K.

Proof.

Simply set H equal to the quotient groupG/K and define Φ:G→G/K to be the natural projection from G to G/K (i.e. Φ sends g∈G to the coset g⁢K). Then it is clear that the kernel of Φ is precisely formed by those elements of K.
∎

Although the first lemma is very simple, it is very useful when one tries to prove that a subgroup is normal.